Displacement, velocity, acceleration. Three key attributes of particles in motion you are no doubt familiar with. What ties them all together though? How are they related mathematically? The answer is calculus.
Modeling Particle Motion Definition
What exactly is modeling particle motion?
Modeling particle motion is the process of describing the motion of a discrete object or 'particle' using mathematical language.
This can be done in several ways, such as with differential equations or transfer functions, but in AP maths it will be done using functions of time.
Modeling Particle Motion as a Function of Time
When modeling a particle's motion as a function of time, you must consider the two variables of the function: the dependent variable, and the independent variable.
Time is always the independent variable, but what is the dependent variable?
Well, that depends on what part of the particle's motion you are modeling: position, displacement, velocity, or acceleration. What might a particle's displacement as a function of time look like? Well...
\[s(t) = t^3 + 2t^2 + t + 3\]
And its velocity?
\[v(t) = 3t^2 + 4t + 1 \]
And its acceleration?
\[a(t) = 6t + 4\]
These examples are polynomials, but conceivably any continuous function of time could model the motion of a particle.
Using these functions, you can calculate the displacement, velocity, and acceleration of a particle at any particular moment in time.
For instance, what is the displacement of the particle at \(3\) seconds? Well...
\[ \begin{align} s(t) &= t^3 + 2t^2 + t + 3 \\\\ s(3) &= 3^3 + 2\cdot 3^2 + 3 + 3 \\\\ &= 27 + 18 + 6 \\\\ &= 51 \,m \end{align} \]
Modeling Particle Motion With Diagrams
To fully understand modeling particle motion, it can be helpful to visualize with diagrams. When modeling particle motion as a function of time in AP math, generally the particle is moving along a straight line, so can be as a number line.
The particle's displacement is defined as its distance from some specified \(0\) point. The particle's displacement is a vector value, meaning it has both size and direction.
The diagram below shows how to measure a particle's position along a number line. The particle's position is measured from a specified \(0\) point.
Fig. 1. Particle Position.
The displacement is similar to the position but subtly different. Displacement describes where the particle is relative to its initial position. We can find the displacement by subtracting the initial position from the current position. If the initial position of the particle is zero then the displacement and position are equivalent.
For instance, if the initial position is \(-1\), and the current position is \(2\), then the displacement is
\[s = 2 - (-1) = 3\]
Fig. 2. Particle Displacement.
The velocity of a particle is the rate of change of its displacement/position. In other words, how quickly the particle's displacement/position is changing. If a particle has a positive velocity, it's moving in the positive direction on the number line (to the right), if it has a negative velocity it is moving in the negative direction of the number line (to the left).
Fig. 3. Particle Velocity.
The acceleration of a particle is the rate of change of its velocity. In other words, how quickly its velocity is changing. If a particle has a positive acceleration, its velocity is increasing in the positive direction, and if a particle has a negative acceleration, its velocity is increasing in the negative direction.
Fig. 4. Particle Acceleration.
It is important to remember that a particle's acceleration and velocity won't necessarily have the same direction!
Right-positive left-negative is not a hard and fast rule, it is just an accepted practice known as a convention. Individual scenarios might set their own convention. But if the question doesn't specify, you should assume that the positive direction goes right. If the line of the particle's movement is vertical, then the positive direction is up.
Modeling Particle Displacement, Velocity, and Acceleration
The definition of velocity is the change in displacement or position with respect to time, in other words, it is the rate of change of displacement or position.
The definition of acceleration is the change in velocity with respect to time, in other words, it is the rate of change of velocity.
Any idea how calculus links these properties?
Well, the way to find the rate of change of a particle, is simply to find its derivative with respect to time. i.e...
\[Displacement \xrightarrow{\frac{d}{dt}} Velocity \xrightarrow{\frac{d}{dt}} Acceleration\]
Yes, that's right, acceleration is the derivative of velocity with respect to time, and velocity is the derivative of displacement with respect to time!
This also means that displacement is the integral of velocity with respect to time, and velocity is the integral of acceleration with respect to time.
\[Displacement \xleftarrow{\int dt + x_0} Velocity \xleftarrow{\int dt + v_0} Acceleration\]
When integrating acceleration, the constant of integration is the initial velocity. When integrating velocity, the constant of integration is the initial position.
Let's look into a couple of examples.
(1) A particle's displacement over time is modeled as
\[x(t) = t^2 + t\]
Find the particle's velocity and acceleration as a function of time.
Solution:
The particle's velocity is the derivative of its displacement with respect to time. Therefore
\[v(t) = 2t + 1 \]
The particle's acceleration is the derivative of its velocity with respect to time. Therefore
\[a(t) = 2\]
(2) A particle's acceleration over time is modeled as
\[a(t) = t \]
Find the particle's velocity and displacement as a function of time, given the initial conditions \(v_0 = 1 \,m/s\) and \(x_0 = 2 \, m \).
Solution:
The particle's velocity is the integral of its acceleration with respect to time. Therefore
\[\begin{align} v(t) &= \int a(t) \, dt + C \\\\ &= \int t \, dt + v_0 \\\\ &= \frac{1}{2} t^2 + 1 \end{align} \]
The particle's displacement is the integral of its velocity with respect to time. Therefore
\[\begin{align} s(t) &= \int v(t) \, dt + C \\\\ &= \int \frac{1}{2}t^2 + 1 \, dt + x_0 \\\\ &= \frac{1}{6} t^3 + t + 2 \end{align} \]
Modeling Particle Motion Examples
Let's look into a few more examples of Modeling Particle Motion.
(1) A particle's straight-line displacement, in meters, can be modeled with the following polynomial.
\[s(t) = x^2 + 2x + 3\]
(a) What is the function of the particle's velocity in terms of time?
(b) What is the particle's acceleration?
(c) Is the particle's velocity increasing or decreasing?
Solution:
(a) The particle's velocity is just the derivative of its displacement with respect to time.
\[\begin{align} v(t) &= \frac{d s}{dt} \\\\ v(t) &= 2x + 2 \end{align} \]
(b) The particle's acceleration is the derivative of its velocity with respect to time.
\[\begin{align} a(t) &= \frac{dv}{dt} \\\\ a(t) &= 2 \ \end{align}\]
(c) The particle's velocity is increasing as it has a positive acceleration.
(4) The graph below shows the velocity of a particle over time. Its initial position is \(x = 2\,m\).
Fig. 5. Graph of particle's velocity over time.
(a) What is the total displacement of the particle after \(6\,s\)?
(b) What is the particle's position after \(4\,s\)?
Solution:
(a) The displacement of the velocity is its integral with respect to time, unfortunately, the question does not supply the velocity as a function in terms of time which can be integrated.
When computing a definite integral, what physical quantity is being solved for? The answer is the area under the graph within the given interval.
In this case, the interval is \(0\,s\) to \(6\,s\). So finding the displacement is a simple case of finding the area under the graph between these two points in time.
\[\begin{align} A &= 2\cdot 2 + 3 \cdot 2 + 1 \cdot 2 \\\\ &= 4 + 6 + 2 \\\\ &= 12 \end{align}\]
Therefore, the total displacement of the particle is \(12\,m\).
(b) Similarly to part (a), the area under the graph in the interval \(0\,s\) to \(4\,s\) is the particle's displacement.
\[\begin{align} A &= 2\cdot 2 + 3 \cdot 2 \\\\ &= 4 + 6 \\\\ &= 10 \end{align}\]
The displacement of the particle is therefore \(10\,m\).
Given that the particle begins at position \(x=2\,m\), its position at \(4\,s\) is
\[\begin{align} x&=s-x_0 \\\\ &= 10-2 \\\\ &=8 \,m \end{align} \]
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